Tutorial: Derivatives of logarithmic and exponential functions
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This tutorial: Part A: Derivatives of logarithmic functions
(This topic is also in Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus)
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#[Derivative of][Derivada de]# ln x #[and][y]# log_{b}x
The derivative of $\ln x$ and $\log_b x$ are given by the following formulas
1. Derivative of the natural logarithm
%%Q Where do these formulas come from?
$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ \gap[20]
2. Derivative of the logarithm with base $b$
$\dfrac{d}{dx}(\log_bx) = \dfrac{1}{x\ln b}$ \gap[5] \t #[When $b = e$, this agrees with (1); see below.][Cuando $b = e$, entonces esto concuerda con (1); vea abajo.]#
#[Note][Nota]# When $b = e$ in formula (2), then $\ln b = \ln e = 1$, so (2) gives the same formula as (1).
%%Examples
1. $\dfrac{d}{dx}(\log_5x) = \dfrac{1}{x\ln\,5}$ \gap[5] \t
Some for you
#[Formula][Fórmula]# 2
\\ 2. $\dfrac{d}{dx}(5\,\ln x) = 5\cdot\dfrac{1}{x} = \dfrac{5}{x}$ \gap[5] \t Derivative of a constant times a function
\\ 3. $\dfrac{d}{dx}(4\,\log_{20} x) = 4\cdot\dfrac{1}{x\ln\,20}$ \gap[5] \t Derivative of a constant times a function
\\ $\qquad {}= \dfrac{4}{x\ln\,20}$
\\ 4. $\dfrac{d}{dx}\left(x^2\,\ln x\right) = 2x\,\ln x+ x^2\dfrac{1}{x}$ \gap[5] \t Product rule
\\ $\qquad {}= 2x\,\ln x+x$
%%A For derivations of these formulas, consult Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus.
Derivatives of logarithms of functions
We now know how to differentiate expressions that contain the logarithm of $x$ with any base. What about the logarithm of a more complicated quantities, for instance $\ln(x^23x+2)$? For things like this we need to use the chain rule (see the %%chainruletut):
Differentiating logarithms of functions
\\ \t
Derivative of the natrual logarithm of a quantity is 1 over that quantity times the derivative of that quantity.
\\
\t The derivative of the log to base b of a quantity is 1 over the product of ln b and that quantity, times the derivative of that quantity.
\\
%%Examples
5. $\dfrac{d}{dx}\ln (\color{blue}{x^2+x}) = \dfrac{1}{\color{blue}{x^2+x}}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= \dfrac{2x+1}{x^2+2}$
\\ 6. $\dfrac{d}{dx}\log_3 (\color{blue}{x^2+x}) = \dfrac{1}{(\color{blue}{x^2+x})\ln 3}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= \dfrac{2x+1}{(x^2+2)\ln 3}$
Some for you
Logs of absolute values
Something curious happens if we take the derivative of the logarithm of the absolute value of $x$:
$\dfrac{d}{dx} \ln \bigx\big$ \t ${}=\dfrac{1}{\bigx\big} \dfrac{d}{dx}\bigx\big$ \t #[Chain rule][Regla de la cadena]#
\\ \t ${}=\dfrac{1}{\bigx\big} \dfrac{\bigx\big}{x}$ \t #[The derivative of $\bigx\big$ is $\frac{\bigx\big}{x}$.][La derivada de $\bigx\big$ es $\frac{\bigx\big}{x}$.]#
\\ \t ${}= \dfrac{1}{x}$ \t Exactly the same as the derivative of $\ln x$!
In other words, replacing $x$ by the absolute value of $x$ has absolutely no effect on the derivative of the natural logarithm. Similarly it has no effect on the derivative of the logarithm of $x$ to any base, or, using the chain rule, on the logarithm of any quantity.
Derivatives of logarithms of absolute values
When taking the derivative of the logarithm of an absolute value, just pretend the absolute value signs were just parentheses, so that the result has no absolute value signs:
When taking the derivative of the logarithm of an absolute value, just pretend the absolute value signs were just parentheses, so that the result has no absolute value signs:
\t
The derivative of the natrual logarithm of the absolute value of a quantity is 1 over that quantity times the derivative of that quantity.
\\
\t The derivative of the log to base b of the absolute value of a quantity is 1 over the product of ln b and that quantity, times the derivative of that quantity.
\\
%%Examples
7. $\dfrac{d}{dx}(\log_5\bigx\big) = \dfrac{1}{x\ln\,5}$ \gap[5] \t
Some for you
#[Compare Example][Comparar el ejemplo]# 1 #[above.][arriba.]#
\\ 8. $\dfrac{d}{dx}(5\,\ln \bigx\big) = 5\cdot\dfrac{1}{x} = \dfrac{5}{x}$ \gap[5] \t #[Compare Example][Comparar el ejemplo]# 2 #[above.][arriba.]#
\\ 9. $\dfrac{d}{dx}\left(x^2\,\ln \bigx\big\right) = 2x\,\ln x+ x^2\dfrac{1}{x}$ \gap[5] \t Compare Example 4 above.
\\ $\qquad {}= 2x\,\ln x+x$
\\ 10. $\dfrac{d}{dx}\ln \big\color{blue}{x^2+x}\big = \dfrac{1}{\color{blue}{x^2+x}}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \gap[5] \t Compare Example 5 above.
\\ $\qquad {}= \dfrac{2x+1}{x^2+2}$
\\ 11. $\dfrac{d}{dx}\log_3 \big\color{blue}{x^2+x}\big = \dfrac{1}{(\color{blue}{x^2+x})\ln 3}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \gap[5] \t Compare Example 6 above.
\\ $\qquad {}= \dfrac{2x+1}{(x^2+2)\ln 3}$
\\ 12. $\dfrac{d}{dx}\ln \big3x+1\big^5 = \dfrac{d}{dx}5\ln \big3x+1\big$ \gap[5] \t Properties of logarithms
\\ $\qquad {}= 5\dfrac{1}{3x+1}\cdot 3 = \dfrac{15}{3x+1}$ \t Compare Example 10 above.
Now try the exercises in Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus.
Copyright © 2019 Stefan Waner and Steven R. Costenoble